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anonymous

  • 4 years ago

find the domain of the function f(x)= 1/sqrt (4x-8) please show calculations

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  1. anonymous
    • 4 years ago
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    huh

  2. anonymous
    • 4 years ago
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    callum can you show the steps please

  3. anonymous
    • 4 years ago
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    (2, infinity)

  4. anonymous
    • 4 years ago
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    \[f(x) = \frac{1}{\sqrt{4x-8}}\] now you want to look at conditions on 4x-8. What can x reasonably be? Well you cannot take a square root of a negative number, and the denominator cannot be zero, so we need: \[4x-8 > 0\] \[x>\frac{1}{2}\] So I was wrong above, the domain is actually: \[\left(\frac{1}{2}, \infty \right)\]

  5. anonymous
    • 4 years ago
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    x > 2

  6. anonymous
    • 4 years ago
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    think you divided the wrong side there

  7. anonymous
    • 4 years ago
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    oh yes sorry it's 2 lol, quite right. \[(2,\infty)\]

  8. anonymous
    • 4 years ago
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    rickjbr can you demonstrate how you got your answer please

  9. anonymous
    • 4 years ago
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    you'll have imaginary numbers if the square root is allowed to go negative

  10. anonymous
    • 4 years ago
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    thus not part of the real number system

  11. anonymous
    • 4 years ago
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    callum was right in his calculation 4x - 8 > 0 4(x-2) > 0 x-2 > 0 x > 2

  12. anonymous
    • 4 years ago
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    yeah my reasoning was right, it's just I divided the inequality wrong and got x>1/2. rickjbr corrected this to x>2

  13. anonymous
    • 4 years ago
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    thanks for the help to both of your

  14. anonymous
    • 4 years ago
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    np

  15. anonymous
    • 4 years ago
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    how about this. 2x^2-3x+5...find and simplify the difference quotient

  16. anonymous
    • 4 years ago
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    The question reads Find and simplify the difference quotient f(x+h)-f(x)/h for the given function f(x)=2x^(2)-3x+5

  17. anonymous
    • 4 years ago
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    ah, right

  18. anonymous
    • 4 years ago
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    (2(x+h)^2-3(x+h)+5-2x^2+3x-5)/h

  19. anonymous
    • 4 years ago
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    (2(x^2+2xh+h^2)-3(x+h)+5-2x^2+3x-5)/h

  20. anonymous
    • 4 years ago
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    (2x^2+4xh+2h^2-3x-3h+5-2x^2+3x-5)/h

  21. anonymous
    • 4 years ago
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    (2h^2+4xh+3x)/h

  22. anonymous
    • 4 years ago
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    is that the final answer

  23. anonymous
    • 4 years ago
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    checking, a bit rusty

  24. anonymous
    • 4 years ago
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    my options are a)4x+2h, b)4x+2h-3, c)4h+2x-3, d)4h+2x, e)4x+4h

  25. anonymous
    • 4 years ago
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    i'll redo it, might've missed something

  26. anonymous
    • 4 years ago
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    ok

  27. anonymous
    • 4 years ago
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    ah, silly

  28. anonymous
    • 4 years ago
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    that last step was it. instead of (2h^2+4xh+3x)/h it should be (2h^2+4xh-3h)/h

  29. anonymous
    • 4 years ago
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    simply factor out an h

  30. anonymous
    • 4 years ago
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    [h(2h+4x-3)]/h =2h+4x-3

  31. anonymous
    • 4 years ago
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    that's it

  32. anonymous
    • 4 years ago
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    thanks greatly. i dont know if you have the time...but im reviewing for a test and I have a whole lot of other questions. if you can't help me, i understand, i will just post on the main forum

  33. anonymous
    • 4 years ago
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    probably better off posting on main forum, if i can help, i'll try

  34. anonymous
    • 4 years ago
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    ok thanks for the help

  35. anonymous
    • 4 years ago
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    Find \[\frac{f(x+h)-f(x)}{h}\] for \[f(x)=2x^2-3x+5\] \[\frac{f(x+h)-f(x)}{h} = \frac{(2(x+h)^2-3(x+h)+5) - (2x^2-3x+5)}{h}\] \[= \frac{(2x^2+4xh+2h^2-3x-3h)-(2x^2-3x)}{h}\] \[=\frac{4xh+2h^2-3h}{h} = 4x+2h-3\] We can verify that this is right by noting that \[f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}\] The left hand side of this is equal to the derivative of f, which we can just do right here: \[f'(x) = 4x - 3\] and we should get that this equals the limit as h goes to zero of \[\frac{f(x+h) - f(x)}{h}.\] We have already calculated that \[\frac{f(x+h) - f(x)}{h} = 4x+2h-3\] so taking the limit of this as h goes to zero we get: \[\lim_{h\rightarrow 0}(4x+2h - 3) = 4x+0-3 = 4x-3.\] Hence we ave verified that the simplification has been done correctly.

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